JEE Mains · Physics · STD 12 - 12. atoms
In a hydrogen spectrum, \(\lambda\) be the wavelength of first transition line of Lyman series. The wavelength difference will be "a \(\lambda^{\prime}\) between the wavelength of \(3^{\text {rd }}\) transition line of Paschen series and that of \(2^{\text {nd }}\) transition line of Balmer Series where \(a=........\)
- A \(5\)
- B \(50\)
- C \(25\)
- D \(4\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
For first line of Lyman \(\frac{1}{\lambda}= R \left(1-\frac{1}{4}\right)= R \left(\frac{3}{4}\right)\) \(\quad \lambda =\frac{4}{3 R } \ldots(1)\) \(3^{\text {rd }} \text { line(Paschen) }\)…
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